Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
int n, m, q, pa[N][17], h[N];
vector<int> c[N], p[N], l[N][17];
vector<int> combine(vector<int> a, vector<int> b) {
  vector<int> ans(a.size() + b.size());
  merge(a.begin(), a.end(), b.begin(), b.end(), ans.begin());
  while (ans.size() > 10) ans.pop_back();
  return ans;
}
void dfs(int par, int u) {
  h[u] = h[par] + 1;
  l[u][0] = p[par];
  pa[u][0] = par;
  for (int i = 0; i < c[u].size(); i++) {
    int v = c[u][i];
    if (v == par) continue;
    dfs(u, v);
  }
}
void rmq() {
  for (int j = 1; j <= 16; j++)
    for (int i = 1; i <= n; i++) {
      if (!pa[i][j - 1]) continue;
      pa[i][j] = pa[pa[i][j - 1]][j - 1];
      l[i][j] = combine(l[i][j - 1], l[pa[i][j - 1]][j - 1]);
    }
}
vector<int> lca(int u, int v) {
  vector<int> ans;
  if (h[u] < h[v]) swap(u, v);
  int del = h[u] - h[v];
  ans = p[u];
  for (int i = 16; i >= 0; i--)
    if ((del >> i) & 1) {
      ans = combine(ans, l[u][i]);
      u = pa[u][i];
    }
  if (u == v) return ans;
  ans = combine(ans, p[v]);
  for (int i = 16; i >= 0; i--)
    if (pa[u][i] != pa[v][i]) {
      ans = combine(ans, l[u][i]);
      ans = combine(ans, l[v][i]);
      u = pa[u][i];
      v = pa[v][i];
    }
  return combine(ans, p[pa[u][0]]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m >> q;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    c[u].push_back(v);
    c[v].push_back(u);
  }
  for (int i = 1, x; i <= m; i++) {
    cin >> x;
    if (p[x].size() == 10) continue;
    p[x].push_back(i);
  }
  dfs(0, 1);
  rmq();
  for (int i = 1; i <= q; i++) {
    int u, v, a;
    cin >> u >> v >> a;
    vector<int> ans = lca(u, v);
    int k = min(a, int(ans.size()));
    cout << k << " ";
    for (int i = 0; i < k; i++) cout << ans[i] << " ";
    cout << endl;
  }
  return 0;
}